A car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s2. how far does the car travel while accelerating?

Science assessment help pls

DENIUS [597]

Answer:

Work Done = W

force = F

Distance = d

W = Fd

or W = F*d

W (in joules) = 3.5*4 = 14 Nm (or J)

1Nm = 1J

so newton meters and joules are the same

Power = Work (in joules) /time (in seconds)

i don’t know the time so i can’t solve it

3 0

3 years ago

Read 2 more answers

What process builds organic molecules such as sucrose by taking water away?

AlexFokin [52]

Dehydration? I think that’s it.

4 0

3 years ago

Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.

bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

$f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}$

3 0

3 years ago

A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconne

Tom [10]

Answer:

Explanation:

Initial separation of plate = d

final separation = 2d

The capacitance of the capacitor will reduce from C to C/2 because

capacitance = ε A / d

d is distance between plates.

As the batteries are disconnected , charge on the capacitor becomes fixed .

Initial charge on the capacitor

= Capacitance x potential difference

Q = C ΔV

Final charge will remain unchanged

Final charge = C ΔV

Final capacitance = C/2

Final potential difference = charge / capacitance

= C ΔV / C/2

= 2 ΔV

Potential difference is doubled after the pates are further separated.

6 0

3 years ago

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

3 years ago